}

> Hello,

> I have a two sets OLDLIST and REMOVE.

> I would like to remove every element in OLDLIST if it is also occuring
> in REMOVE
> and store the remaining elements from OLDLIST into NEWLIST.

> So far, I have read in both lists but I'm struggling comparing the
> elements
> - I have something like

> double oldlist[oldsize];
> double remove[remsize];

> for(i = 0; i < oldsize; i++)
> {

>     for(k = 0; k < remsize; k ++)

>        if(oldlist[i] != remove[k])
>           {
>               cout<<oldlist[k]<<endl;

>           }

> }

Sean Dalton <sean.dalton...@yahoo.ca> wrote:
> I have a two sets OLDLIST and REMOVE.

> I would like to remove every element in OLDLIST if it is also
> occuring in REMOVE and store the remaining elements from OLDLIST
> into NEWLIST.

> On Jul 5, 4:43 pm, Sean Dalton <sean.dalton...@yahoo.ca> wrote:
> > I have a two sets OLDLIST and REMOVE.
> > I would like to remove every element in OLDLIST if it is
> > also occuring in REMOVE and store the remaining elements
> > from OLDLIST into NEWLIST.
> > So far, I have read in both lists but I'm struggling
> > comparing the elements
> > - I have something like
> > double oldlist[oldsize];
> > double remove[remsize];
> > for(i = 0; i < oldsize; i++)
> > {
> >     for(k = 0; k < remsize; k ++)
> >        if(oldlist[i] != remove[k])
> >           {
> >               cout<<oldlist[k]<<endl;
> >           }
> > }
> Are you sure you want to use arrays to store your data?

> This would be a n-squared algorithm.

> Better would be to store the remove set into a hash_map or
> some other container which can do a constant time lookup to
> see if you need to remove the data.